# Proving on the equation $\frac1{x+1}+\frac1{2(x+2)}+\frac1{3(x+3)}+\cdots+\frac1{n(x+n)}=1$

Consider the equation: $$\begin{gather*} \frac{1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + \cdots + \frac{1} {n(x+n)} = 1.\tag1 \end{gather*}$$ Prove that:

• For every $$n \geq 2$$, equation $$(1)$$ always has a unique positive solution $$x_n$$.

• The sequence $$(x_n)$$ has a finite limit with $$n \to \infty+$$. Find that limit.

Here's all I did:

$$\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)} = 1.$$

$$\Leftrightarrow 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) = n! (x+1)(x+2)...(x+n)$$

Consider the polynomial $$P(x) = 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) - n! (x+1)(x+2)...(x+n)$$

Consider the highest coefficient of the polynomial:$$-n! < 0$$

Consider the lowest coefficient of the polynomial:$$(2.3...n)^2+(1.3...n)^2+...(1.2...(n-1))^2 - n! = (1.3...n)^2+...(1.2...(n-1))^2 > 0$$

Let $$x_1,x_2,x_3....$$ be the solutions of$$P(x)$$

Assume that equation (1) has no positive solution.

According to Viette's theorem: $$x_1 x_2 x_3....x_n = (-1)^n \frac{a_0}{a_n}$$

$$\frac {a_0}{a_n} < 0 \Rightarrow$$if $$n$$ is an odd number , then $$x_1 x_2 x_3....x_n < 0$$ , but

$$(-1)^n \frac{a_0}{a_n} >0 \Rightarrow$$ (nonsense) . Same with $$n$$being an even number.

So the polynomial must have at least one integer root.

Suppose the polynomial has at least 2 positive roots $$x$$ and $$y$$ .

$$\frac {1}{y+1} + \frac{1}{2(y+2)} + \frac{1}{3(y+3)} + .... + \frac {1} {n(y+n)} =\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)} .$$

$$\Rightarrow (y-x)( \frac {1}{(x+1)(y+1)}+ \frac {1}{2(x+2)(y+2)} +...+\frac {1}{n(x+n)(y+n)} ) = 0$$

$$\Rightarrow x=y \Rightarrow$$ (no sense)

So for every $$n \geq 2$$ , equation $$( 1 )$$ always has a unique positive solution $$x_n$$

That's all I did , and I have no idea what to do next , I want to establish a relationship between $$x_n$$ but for $$n \geq 3$$ the polynomial's solution is a " bad " solution . " so it's very difficult to build. Hope to get help from everyone. Thanks very much .

• "polynomial must have at least one integer root" should perhaps be "polynomial must have at least one positive root" Jul 21, 2021 at 10:55

Let $$f_n(x) = \frac{1}{x+1} + \cdots + \frac{1}{n(x+n)}.$$

Observe that $$f_n(x)$$ is strictly decreasing in $$x$$ for $$x\ge 0$$. Further, for $$n\ge 2$$, $$f_n(0) > 1$$ and $$\lim_{x\to\infty} f_n(x) = 0$$. It follows that there is a unique positive solution $$x_n$$ to $$f_n(x) = 1$$.

Let $$f$$ be defined by

$$f(x) = \sum_{i=1}^\infty \frac{1}{i(x+i)}.$$ Note that $$(f_n)$$ uniformly converges to $$f$$ (for $$x\ge 0$$, $$|(f-f_N)(x)| \le \sum_{i=N}^\infty 1/i^2$$, which can be made arbitrarily small for sufficiently large $$N$$).
Let the $$x_n$$ converge to $$l$$. Then, uniform convergence implies that $$1 = \lim_{n\to\infty} f_n(x_n) = f(l)$$. So,

\begin{align*} 1 &= \sum_{i=1}^\infty \frac{1}{i(l+i)} \\ l &= \sum_{i=1}^\infty \frac{1}{i} - \frac{1}{l+i}. \end{align*} Equality is seen to hold for $$l=1$$ since then, the right hand side becomes a telescoping sum which is evaluated to be $$1$$, and the left is $$1$$.

Thus, $$x_n \to 1$$.

• Whar does " uniformly converges " mean ? I have absolutely no idea about this, can you help me? Jul 21, 2021 at 14:18
• – Gary
Jul 21, 2021 at 14:23

Since you already find a solution for the first question. I write an answer for the second one. Observe that $$\sum_{k=1}^n\frac{1}{k\cdot (k+1)}=\sum_{k=1}^n\frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{n+1}.$$ Hence $$\sum_{k=1}^{\infty}\frac{1}{k\cdot (k+1)}=1.$$ This means for $$x=1$$ the left handside of your equation tends to 1. One can easily prove, that for $$x<1$$ and sufficient large $$n$$ the lefthandside is greater than 1 and for $$x>1$$ and sufficient large $$n$$ the lefthandside is less than 1.

For you first question, a picture may help clarify what is happening. This one comes from Wolfram Alpha when $$n=7$$

In effect your function, let's call it $$f_n(x)$$, is the sum of $$n$$ hyperbolae with $$n$$ vertical asymptotes at $$-1,-2,\ldots, -n$$, a locally decreasing function of $$x$$ where it exists, and with $$f_n(x) \to 0^-$$ as $$x \to - \infty$$ and $$f_n(x) \to 0^+$$ as $$x \to + \infty$$. So $$f_n(x)=1$$ gives $$n$$ real solutions, of which $$n-1$$ are clearly negative and, since $$f_n(0)\ge 1$$, one must be non-negative and when $$n>1$$ is positive.

As others have said the limit as $$n\to \infty$$ comes when $$x=1$$ and you get $$\sum\limits_{k=1}^\infty \frac{1}{k(k+1)}=1$$